Question

If the coefficients of $5^{th}$,$6^{th}$ and $7^{th}$ terms in the expansion of ${(1+x)}^n$ are in $A.P.$, find $n$.

Answer 1

$(1+x{)}^n=1+{}^n{C}_{1}x+{}^n{C}_2{x}^2+{}^n{C}_3{x}^3+{}^n{C}_4{x}^4+{}^n{C}_5{x}^5+{}^n{C}_6{x}^6+...$

$5^{th}$ term= ${}^n{C}_4$

$6^{th}$ term= ${}^n{C}_5$

$7^{th}$ term= ${}^n{C}_6$

$T_5,T_6,T_7$ are in AP i.e. $2T_6=T_5+T_7$

Putting values of $T_5,T_6,T_7$ in the equation

$2{}^n{C}_5={}^n{C}_4+{}^n{C}_{6}2\frac{n(n-1)(n-2)(n-3)(n-4)}{120}=\frac{n(n-1)(n-2)(n-3)}{4!}+\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{6!}\frac{n(n-1)(n-2)(n-3)(n-4)}{60}=\frac{n(n-1)(n-2)(n-3)}{4!}[1-\frac{(n-4)(n-5)}{30}]\frac{n(n-1)(n-2)(n-3)(n-4)}{60}=\frac{n(n-1)(n-2)(n-3)}{4!}\frac{(30-n^2+9n-20)}{30}\frac{n(n-1)(n-2)(n-3)(n-4)}{60}=\frac{-n(n-1)(n-2)(n-3)}{6!}(n^2-9n-10)\frac{n(n-1)(n-2)(n-3)(n-4)}{60}=\frac{-n(n-1)(n-2)(n-3)(n-10)(n+1)}{6!}(n-4)=\frac{-(n-10)(n+1)}{2}\Rightarrow n^2-9n-10+2n-8=0n^2-7n-18=0(n-9)(n+2)=0n=9$